The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $20.5$ years; the standard deviation is $3.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living longer than $10$ years.
Solution: $20.5$ $17$ $24$ $13.5$ $27.5$ $10$ $31$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $20.5$ years. We know the standard deviation is $3.5$ years, so one standard deviation below the mean is $17$ years and one standard deviation above the mean is $24$ years. Two standard deviations below the mean is $13.5$ years and two standard deviations above the mean is $27.5$ years. Three standard deviations below the mean is $10$ years and three standard deviations above the mean is $31$ years. We are interested in the probability of a zebra living longer than $10$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the zebras will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $10$ years and the other half $({0.15\%})$ will live longer than $31$ years. The probability of a particular zebra living longer than $10$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.